JavaScript 曲线生成

Question

如何编写一个函数，接受 2D 点数组并返回贝塞尔/二次曲线，以便稍后使用 HTML5 Canvas bezierCurveTo 或 quadraticCurveTo 重新绘制它方法？

Answer 1

编辑：改进。

查看使用以下代码的演示。

var makeCurveArgs = function(points) {
  var copy = points.slice();
  var result = [];
  copy.shift(); //drop the first point, it will be handled elsewhere
  var tangent;
  if(copy.length >= 3) {
    var cp1 = copy.shift();
    var cp2 = copy.shift();
    var p2 = copy.shift();

    result.push([cp1[0], cp1[1], cp2[0], cp2[1], p2[0], p2[1]]);            
  }
  while(copy.length >= 2) {
    var cp1 = [2 * p2[0] - cp2[0], 2 * p2[1] - cp2[1]];
    var cp2 = copy.shift();
    var p2 = copy.shift();
    result.push([cp1[0], cp1[1], cp2[0], cp2[1], p2[0], p2[1]]);            
  }
  return result;
}

var notThatHard = function(points) {
  var origin = points[0].slice();
  var curves = makeCurveArgs(points);
  var drawCurves = function(context) {


    context.beginPath();
    context.moveTo(origin[0], origin[1]);
      for(var i = 0; i < curves.length; i++) {
      var c = curves[i];
      context.bezierCurveTo(c[0], c[1], c[2], c[3], c[4], c[5]);
  }
  };
  return drawCurves;
};

一般方法是，您给我点和控制点的坐标，然后我给您返回一个函数，该函数将在画布上下文上执行该路径。

我给出的函数需要一个 2N+2 2 元素数组的数组；每个 2 元素数组都是一个 (x,y) 坐标。坐标的使用方式如下：

points[0]: starting point of the curve
points[1]: lies on a line tangent to the beginning of the 1st bezier curve
points[2]: lies on a line tangent to the end of the 1st bezier curve
points[3]: end of 1st bezier curve, start of 2nd bezier curve
points[4]: lies on a line tangent to the end of the 2nd bezier curve
points[5]: end of 2nd bezier curve, start of 3rd curve
...
points[2*K+2]: lies on a line tangent to the end of the Kth bezier curve
points[2*K+3]: end of Kth bezier curve, start of (K+1)th

我认为编写一个类似的 QuadraticCurveTo 函数并不难。